$\begin{cases}c(1)=-20\\\\ c(n)=c(n-1)+10 \end{cases}$ Find the $2^{\text{nd}}$ term in the sequence.
Explanation: This is a recursive formula. It tells us that the first term is $-20$ and that the common difference is $10$. $\begin{aligned} {c(1)}&=-20 \\\\ {c(2)}&={c(1)}+10=-10 \end{aligned}$ The $2^{\text{nd}}$ term is $-10$.